Some Basic Concepts of Chemistry

1.25. Molar mass of Na₂CO₃= (2 x 23) + 12 + (3 x 16) = 106g mol^-1 

0.50 mol Na₂CO₃ means 0.50 x 106 g = 53 g of Na₂CO₃

0.50 M Na₂CO₃ means 0.50 mol per volume in litre,

i.e., half of 106 g Na₂CO₃ is present in 1 L solution.

i.e.,53 g Na₂CO₃ is present in 1 L of the solution

1.24. According to the given equation, 1 mol of N₂ reacts with 3 mol of H₂.

Or, 28 g of N₂ react with 6 g of H₂.

So, 2000 g of N₂ will react with H₂ = 6/28 x 2000 g = 428.6 g of H₂.

(i) 2 mol of N₂ or 28 g of N₂ produce NH₃ = 2 mol = 34 g

So, 2000 g of N₂ will produce NH₃ = 34/28 x 2000 g = 2428.57 g

(ii) Yes, N₂ is the limiting reagent while H₂ is the excess reagent. So, H₂ will remain unreacted.

(iii) H₂ will remain unreacted. Mass left unreacted = 1000 g – 428.6 g = 571.4 g

1.23.  (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
Thus, 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence, B is the limiting reagent while A is the excess reagent.

(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
Thus, 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant since 1 mol of B is left unreacted.

(iii) No limiting reagent.

(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.

(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent

1.22. Speed = Distance / Time

Or, Distance = Speed x Time = 3.0 * 10⁸ms^–1 x 2.00 ns = 3.0 * 10⁸ms^–1 x 2.00 x 10^-9 s = 6 x 10^-1 m = 0.600 m

1.20. The values after round up with three significant figures are:

(i) 34.2

(ii) 10.4

(iii) 0.0460

(iv) 2810

1.19. The significant figures in the given values are:

(i) 2 in 0.0025

(ii) 3 in 208

(iii) 4 in 5005

(iv) 3 in 126,000

(v) 4 in 500.0

(vi) 5 in 2.0034

1.18. The scientific notation of the given values are:

(i) 4.8*10^?3

(ii) 2.34*10⁵

(iii) 8.008*10³

(iv) 5.000*10²

(v) 6.0012*10⁰

1.17. 15 ppm means 15 parts per million i.e.15 in 10⁶

So, % by mass = 15/10⁶ x 100 = 15 x 10^-4 = 1.5 x 10^-3%

Molality = No. of moles of solute/Mass of solvent in kg

Percent by mass = 1.5 x 10^-3 % means 100 g of the sample contain 1.5 x 10^-3 g chloroform.

So, 1000 g or 1 kg of the sample will contain 1.5 x 10^-3 x 1000/100 = 1.5 x 10^-2 g chloroform.

Molar mass of chloroform = 12 + 1 + (3 x 35.5) = 119.5 g/mol

Therefore, molality = 1.5 x 10^-2 /119.5 = 1.26 x 10^-4m.

1.16. Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures